Question
Show graphically that the following system of equation is in-consistent (i.e. has no solution):
3x - 5y = 20
6x - 10y = -40
✓
Answer
We have,
3x - 5y = 20
6x - 10y = -40
Now, 3x - 5y = 20
$\Rightarrow\text{x}=\frac{5\text{y}+20}{3}$
When y = -1, we have,
$\text{x}=\frac{5(-1)+20}{3}=5$
When y = -4, we have,
$\text{x}=\frac{5(-4)+20}{3}=0$
Thus we have the following table giving points on the line 3x - 5y = 20.
Now, 6x - 10y = -40
⇒ 6x = -40 + 10y
$\Rightarrow\text{x}=\frac{-40+10\text{y}}{6}$
When y = 4, we have,
$\text{x}=\frac{-40+10\times4}{6}=0$
When y = 1, we have,
$\text{x}=\frac{-40+10\times1}{6}=-5$
Thus we have the following table giving points on the line 6x - 10y = -40
Graph of the given equations:
Clearly, there is no common points between these two lines.
Hence, given system of equations is in-consistent.
3x - 5y = 20
6x - 10y = -40
Now, 3x - 5y = 20
$\Rightarrow\text{x}=\frac{5\text{y}+20}{3}$
When y = -1, we have,
$\text{x}=\frac{5(-1)+20}{3}=5$
When y = -4, we have,
$\text{x}=\frac{5(-4)+20}{3}=0$
Thus we have the following table giving points on the line 3x - 5y = 20.
|
x
|
5
|
0
|
|
y
|
-1
|
-4
|
⇒ 6x = -40 + 10y
$\Rightarrow\text{x}=\frac{-40+10\text{y}}{6}$
When y = 4, we have,
$\text{x}=\frac{-40+10\times4}{6}=0$
When y = 1, we have,
$\text{x}=\frac{-40+10\times1}{6}=-5$
Thus we have the following table giving points on the line 6x - 10y = -40
|
x
|
0
|
-5
|
|
y
|
4
|
1
|
Clearly, there is no common points between these two lines.
Hence, given system of equations is in-consistent.
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